High School

It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (molecular weight: 46.07 g/mol). What will be [tex]\Delta S_{\text{sys}}[/tex] for the vaporization of 5.00 g of ethanol at 79.8 °C?

Answer :

To find the change in entropy (delta S_{sys}) for the vaporization of 5.00 grams of ethanol, we use the provided enthalpy of vaporization and convert the given amount to moles before calculating delta S_{sys}
using the formula ΔS = q_{rev} / T.

The question is asking to calculate the change in entropy (delta S_{sys}) for the vaporization of 5.00 grams of ethanol at a temperature of 79.8
°C. To find the entropy change for the given amount of ethanol, we use the provided enthalpy of vaporization for 1 mol of ethanol, which is 38.6 kJ. The number of moles of 5.00 g ethanol is calculated using its molar mass (46.07 g/mol). Then, using the formula Enter your alt text here

ΔS = q_{rev} / T, where ΔS is the entropy change, q_{rev} is the heat involved in the reversible process (enthalpy of vaporization in this case), and T is the temperature in Kelvin, we can find delta S_{sys}. The final temperature must be converted to Kelvin by adding 273.15 to the Celsius value.